SOLUTION: At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales rea

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales rea      Log On


   

Question 1179222: At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales reached $3920. If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold, how many tickets of each type were sold?
Found 3 solutions by josgarithmetic, MathLover1, ikleyn:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
PRICES         COUNTS           COSTS
  8              e               8e
 10            420-e-t          10(420-e-t)
 12              t              12t
               420               3920

system%28e%2B420-e-t=5t%2C8e%2B10%28420-e-t%29%2B12t=3920%29

First equation gives highlight%28t=70%29.

Second equation simplifies to e-t=140.
Substitution gives highlight%28e=210%29.

Quantity of $10 tickets by difference, highlight%28140%29.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

At a local Brownsville play production, 420 tickets were sold.
let's ticket that cost $8 be a, ticket that cost $10 be b, and ticket that cost $12 be c

a%2Bb%2Bc=420.........eq.1

if the total income from ticket sales reached $3920, we have
8a%2B10b%2B12c=3920.....eq.2
If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12+tickets, we have

a%2Bb=5c....eq.3

go to
a%2Bb%2Bc=420.........eq.1, substitute a%2Bb from eq.3
5c%2Bc=420
6c=420
c=420%2F6
c=70
go to
a%2Bb=5%2A70....eq.3
a%2Bb=350..........solve for a
a=350-b..............eq.1)

go to
8a%2B10b%2B12c=3920.....eq.2, substitute a and c
8%28350-b%29%2B10b%2B12%2A70=3920.......solve for b
2800-8b%2B10b%2B840=3920
3640%2B2b=3920
2b=3920-3640
2b=280
b=140

go to

a=350-b..............eq.1), substitute b
a=350-140
a=210

sold, how many tickets of each type were sold?
210 tickets that cost $8 were sold,
140tickets that cost $10 were sold,
and 70 tickets that cost $12 were sold

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
At a local Brownsville play production, 420 tickets were sold.
The ticket prices varied on the seating arrangements and cost $8, $10, or $12.
The total income from ticket sales reached $3920.
If the combined number of $8 and $10 priced tickets sold was 5 times the number of $12 tickets sold,
how many tickets of each type were sold?
~~~~~~~~~~~~ 

Let x be the number of the $8 tickets, and

let y be the number of the $10 tickets.


Then, from the condition, the number  (x+y)  is  5/6  of 420, i.e. x+y = 350,

      and the number of those who bought the $12 tickets was the rest  1/6 of 420, i.e. 70 persons.


Now we have the system of 2 (two) equations in two unknowns


     x +   y         =  350     (1)    

    8x + 10y + 12*70 = 3920     (2)    (total revenue)


We simplify this system to this EQUIVALENT strandard form


      x +   y        =  350     (3)  

     8x + 10y        = 3080     (4)  


Multiply equation (3) by 10 (both sides) and then subtract from it equation (4).  You will get then


    10x - 8x         = 3500 - 3080

       2x            =  420

        x            = 420/2 = 210.


Thus 210 persons bought $8 tickets;  hence,  350-210 = 140 bought $10 tickets;  and the rest  70 persons bought $12 tickets.

Solved.


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By the way,  the problem' setup,  presented in the post by @josgarithmetic,

                        IS TOTALLY WRONG,


            so for your safety,  you better ignore it . . .